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Ii. What mass in grams of sodium trioxocarbonate (IV) is needed to prepare 250 cm³ of 0.2 moldm³ solution? [Na = 23, O=16, C=12]

Chemistry

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here are the solutions to the chemistry questions:

4. (a) i. What is a buffer solution?
A buffer solution is a solution that resists significant changes in pH when small amounts of an acid or a base are added. It typically consists of a weak acid and its conjugate base, or a weak base and its conjugate acid.

4. (a) ii. Give two uses of buffer solution.

  • Maintaining the pH of blood and other biological fluids.
  • Controlling the pH in industrial processes, such as fermentation and dyeing.

4. (b) i. Name the process used in the production of ammonia.

  • Haber process

4. (b) ii. Give two physical properties of nitrogen.

  • It is a colorless gas.
  • It is an odorless gas.

4. (b) iii. List the two allotropes of carbon.

  • Diamond
  • Graphite

4. (b) iv. Mention the structure of each allotrope in (b)iii.

  • Diamond: Each carbon atom is covalently bonded to four other carbon atoms in a tetrahedral arrangement, forming a giant covalent network structure.
  • Graphite: Each carbon atom is covalently bonded to three other carbon atoms in a trigonal planar arrangement, forming hexagonal layers. These layers are held together by weak intermolecular forces.

4. (b) v. Give one use of charcoal.

  • Used as a fuel.

4. (c) i. Give the general molecular formula for:
I. Alkenes: CnH2n\boxed{\text{C}_n\text{H}_{2n}}
II. Alkynes: CnH2n−2\boxed{\text{C}_n\text{H}_{2n-2}}

4. (c) ii. Write out the possible structural isomers of C4H10\text{C}_4\text{H}_{10}.
I. CH3CH2CH2CH3\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3
II. CH3CH(CH3)CH3\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_3

4. (c) iii. Give the IUPAC names of the isomers in (c)ii.
I. Butane\boxed{\text{Butane}}
II. 2-Methylpropane\boxed{\text{2-Methylpropane}}

4. (c) iv. Calculate the molecular formula of a compound with the following composition by mass; 60% carbon, 13.3% hydrogen and 20.7% oxygen with a molar mass of 60.
Note: The given percentages (60% C, 13.3% H, 20.7% O) sum to 94%. For the problem to be consistent with a molar mass of 60 g/mol, we assume the oxygen percentage is 100%−60%−13.3%=26.7%100\% - 60\% - 13.3\% = 26.7\%.

  1. Determine the mass of each element in 100 g of the compound.

    Mass of C = 60 g
    Mass of H = 13.3 g
    Mass of O = 26.7 g (assumed)

  2. Convert the mass of each element to moles.

    Moles of C=60 g12 g/mol=5 mol\text{Moles of C} = \frac{60 \text{ g}}{12 \text{ g/mol}} = 5 \text{ mol} Moles of H=13.3 g1 g/mol=13.3 mol\text{Moles of H} = \frac{13.3 \text{ g}}{1 \text{ g/mol}} = 13.3 \text{ mol} Moles of O=26.7 g16 g/mol≈1.66875 mol\text{Moles of O} = \frac{26.7 \text{ g}}{16 \text{ g/mol}} \approx 1.66875 \text{ mol}
  3. Determine the simplest whole-number mole ratio by dividing by the smallest number of moles.

    The smallest number of moles is 1.66875 (for Oxygen).

    Ratio C=51.66875≈2.996≈3\text{Ratio C} = \frac{5}{1.66875} \approx 2.996 \approx 3 Ratio H=13.31.66875≈7.97≈8\text{Ratio H} = \frac{13.3}{1.66875} \approx 7.97 \approx 8 Ratio O=1.668751.66875=1\text{Ratio O} = \frac{1.66875}{1.66875} = 1

    The empirical formula is C3H8O\text{C}_3\text{H}_8\text{O}.

  4. Calculate the empirical formula mass.

    Empirical formula mass=(3×12)+(8×1)+(1×16)=36+8+16=60 g/mol\text{Empirical formula mass} = (3 \times 12) + (8 \times 1) + (1 \times 16) = 36 + 8 + 16 = 60 \text{ g/mol}
  5. Determine the molecular formula.

    The given molar mass is 60 g/mol. Since the empirical formula mass (60 g/mol) is equal to the molar mass (60 g/mol), the molecular formula is the same as the empirical formula.
    The molecular formula is C3H8O\boxed{\text{C}_3\text{H}_8\text{O}}.

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